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### Course: Calculus, all content (2017 edition)>Unit 1

Lesson 15: Limits of trigonometric functions

# Trig limit using Pythagorean identity

In this video, we explore finding the limit as θ approaches 0 for the expression (1-cosθ)/(2sin²θ). By using the Pythagorean identity, we rewrite the expression to simplify it and avoid the indeterminate form 0/0. This allows us to evaluate the limit and find the answer, 1/4.

## Want to join the conversation?

• I'm still having confusion understanding why Sal changes the function from f(x) to g(x) at . Why does he do that?
(47 votes)
• We know that theta won't be equal to zero as we approach it to zero. So why it's necessary to define another function (namely g(x)) to calculate the limit? We aren't calculating the f(theta=0).
(22 votes)
• besides using a cheat sheet and doing a ton of problems, is there a simple way to either remember or derive the double (and half- for that matter) angle identities. you know, something like the magic hexagon or a nice mnemonic?
(17 votes)
• Found this when I was in 10th grade been using it since:

For sin(a+b)= sin(a)cos(b)+cos(a)sin(b)
Say out loud: "sine cosine cosine sine"

For cos(a+b)= cos(a)cos(b)-sin(a)sin(b)
Say out loud: "cosine cosine SIGN sine sine"
Notice the word "SIGN". The word SIGN is there to remind you that the SIGN of the rightmost term has the opposite SIGN of the input to the cosine on the left hand side of the equation (if its a+b, then subtract. if its a-b, then add).

Just say it out loud like 3 times. It has a really nice rythm:

"Sine Cosine Cosine Sine"
"Cosine Cosine SIGN Sine Sine"
"Sine Cosine Cosine Sine"
"Cosine Cosine SIGN Sine Sine"
"Sine Cosine Cosine Sine"
"Cosine Cosine SIGN Sine Sine"
(100 votes)
• At Sal says that when we arrive at 0/0 we can still find the limit for this function, but if there was a different number in the numerator that is divided by zero that we would know that the limit for this function does not exist. Why is it that we know the limit does not exist when a non-zero number is divided by zero, but not when 0/0? Is this just because we can use the methods of rationalization and factorization on some expressions that work out to 0/0 but not on expressions that work out to non-zero/0? Or is there a deeper reason for this?
(22 votes)
• You will notice that many limit problems are in the form 0/0 and the techniques we use to evaluate these are by rationalizing or factoring and canceling. Later on we will learn about L'Hopital's rule as another technique. So we know that when it is 0/0 we know that it may or may not exist and we need to use one of the techniques to find out.

When the numerator is a non-zero, called it a, then a/0 is undefined (dividing by 0 is undefined), therefore its limit doesn't exist. In term of limit, a/0 = ±∞, depends on the sign. When a limit is ±∞, it is called infinite limit or unbound limit. In the infinite limit, the limit doesn't exist because we do not treat ±∞ as a number.
(17 votes)
• At , I'm still a little confused... why can theta not equal zero if you cancel out the ( 1 - cos(theta))?
(8 votes)
• You want the functions to be equivalent to each other after you've simplified.

If you simplify the function by canceling out something in the denominator you'll need to watch out for values that would've made the denominator zero before you canceled them out, otherwise you would "create" a defined function value that wasn't there before.

Likewise if you try to rewrite an arbitrary function in a more complicated way and you create cases, where it is not defined, you'll need to add them somehow to make sure that the function stays exactly the same, here's an example:

if you want to rewrite the function f(x) = 1 in a more complicated way you could do something like f(x) = 1 = (x)^0 = (x)^(1-1) = x^(1) * x^(-1) = x/x but suddenly your function is not defined at x = 0 anymore because you'll get 0/0 (which is undefined since you can't divide by 0) instead of 1 as a result and you would have to add a condition that makes sure that your function is still defined at that point by writing f(x) = x/x for x not equal to 0, and f(x) = 1 for x = 0. (as piecewise defined function for example).

Basically you do this to stay "mathematically rigorous".
(12 votes)
• Why does 0/0 mean indeterminate and not undefined?
(4 votes)
• Suppose 0/0 = x. This implies that 0*x = 0. However, any value of x will satisfy the equation, so it is indeterminate. (This is different from an undefined expression which has no solutions.)
(14 votes)
• When we cancel out the 1-cosx, shouldn't we also say that x cannot equal 180 as cos 180 will be -1 and 2(1+-1)(1-cosx) will make the denominator equal 0
(7 votes)
• Well, in fact x does not equal neither 180° nor 0°, because we're considering limit as x approaches 0, that means it gets infinitely close to 0 but never actually equals it, so the denominator does not equal 0 either.
Hope that helps:)
(3 votes)
• What should I watch to be brought up to speed on trig? I don’t remember anything other than sah cah toa
(6 votes)
• Check out the Trigonometry course on KA, or look up things like sin and cos identities if you need a review.
(4 votes)
• At , could someone please explain why 0/0 would not signify that there is no limit while any other number over 0 would signify that there is no limit?
(3 votes)
• Finding 0/0 means answering the question “what number times 0 is 0?”. Any number times 0 is 0, so any number can equal 0/0. So the value of 0/0 is inconclusive (or indeterminate), meaning that more work needs to be done to find the limit or determine that the limit doesn’t exist.

Let b represent a nonzero number. Finding b/0 means answering the question “what number times 0 is b”. No number times 0 is the nonzero number b, so no number can equal b/0. This means the limit definitely doesn’t exist.
(8 votes)
• at around , why does Sal say theta is not equal to zero? and then why does he change the function to g(x)? can someone please explain this part of the video?
(3 votes)
• There is a removable discontinuity at zero but you can find the limit by ignoring
the factor (1 - cos(theta)) in both the numerator and denominator. He's doing exactly the same you did when you had rational functions with removable discontinuities.
For example if you need the limit as x --> 1 of the function [ (x - 1) (x + 2) ] / [ (x - 1) (x + 3) ] you only need to find the limit as x --> 1 of the function (x + 2) / (x + 3) , which is doable by direct evaluation.
When you want to communicate this properly you give a different name for the two functions, say f for the first and g for the second. The difference between the two: you get in trouble by directly evaluating f at x=1 but you don't with g.
Now in trig the independent variable is frequently theta or some other cap from the greek well - but don't let that throw you. theta is just like x.
In Sal's example f cannot be evaluated directly at theta=0, but he made a new function by
"cancelling" the (1 - cos(theta)) factor from the numerator and denominator.
Hope this helps.
(6 votes)
• I understand 0/0 is an indeterminate form meaning we can find the actual limit using different methods, but how would you know if the limit is actually 0?
(3 votes)
• If at some point in the process of finding the limit, we find that the numerator goes to 0 while the denominator goes to a nonzero number, then the limit is 0.

This is not the only way the limit can be 0. For example, if, at some point in the process of finding the limit, we find that the numerator remains bounded while the denominator goes towards infinity or -infinity, then the limit is 0.
(5 votes)

## Video transcript

- [Voiceover] Let's see if we can find the limit as theta approaches zero of one minus cosine theta over two sine squared theta. And like always, pause the video and see if you could work through this. Alright, well our first temptation is to say, "Well, this is going to be the same thing "as the limit of one minus cosine theta "as X approaches, or not X, "as theta approaches zero. "of theta, as theta approaches zero, "over the limit, "as theta approaches zero "of two sine squared theta." Now, both of these expressions which could be used to define a function, that they'd be continuous if you graph them. They'd be continuous at theta equals zero, so the limit is going to be the same thing, as just evaluating them at theta equals zero So this is going to be equal to one minus cosine of zero over two sine squared of zero. Now, cosine of zero is one and then one minus one is zero, and sine of zero is zero, and you square it, You still got zero and you multiply times two, you still got zero. So you got zero over zero. So once again, we have that indeterminate form. And once again, this indeterminate form when you have zero over zero, doesn't mean to give up, it doesn't mean that the limit doesn't exist. It just means, well maybe there's some other approaches here to work on. If you got some non-zero number divided by zero, then you say, okay that limit doesn't exist and you would say, well you just say it doesn't exist. But let's see what we can do to maybe, to maybe think about this expression in a different way. So if we said, so let's just say that this, let me use some other colors here. Let's say that this right over here is F of X. So, F of X is equal to one minus cosine theta over two sine squared theta, and let's see if we can rewrite it in some way that at least the limit as theta approaches zero isn't going to, we're not gonna get the same zero over zero. Well, we can, we got some trig functions here, so maybe we can use some of our trig identities to simplify this. And the one that jumps out at me is that we have the sine squared of theta and we know from the Pythagorean, Pythagorean Identity in Trigonometry, it comes straight out of the unit circle definition of sine and cosine. We know that, we know that sine squared theta plus cosine squared theta is equal to one or, we know that sine squared theta is one minus cosine squared theta. One minus cosine squared theta. So we could rewrite this. This is equal to one minus cosine theta over two times one minus cosine squared theta. Now, this is one minus cosine theta. This is a one minus cosine squared theta, so it's not completely obvious yet of how you can simplify it, until you realize that this could be viewed as a difference of squares. If you view this as, if you view this as A squared minus B squared, we know that this can be factored as A plus B times A minus B. So I could rewrite this. This is equal to one minus cosine theta over two times, I could write this as one plus cosine theta times one minus cosine theta. One plus cosine theta times one minus, one minus cosine theta. And now this is interesting. I have one minus cosine theta in the numerator and I have a one minus cosine theta in the denominator. Now we might be tempted to say, "Oh, let's just cross that out with that "and we would get, we would simplify it "and get F of X is equal to one over "and we could distribute this two now." We could say, "Two plus two cosine theta." We could say, "Well, aren't these the same thing?" And we would be almost right, because F of X, this one right over here, this, this is defined this right over here is defined when theta is equal to zero, while this one is not defined when theta is equal to zero. When theta is equal to zero, you have a zero in the denominator. And so what we need to do in order for this F of X or in order to be, for this to be the same thing, we have to say, theta cannot be equal to zero. But now let's think about the limit again. Essentially, what we want to do is we want to find the limit as theta approaches zero of F of X. And we can't just do direct substitution into, if we do, if we really take this seriously, 'cause we're gonna like, "Oh well, if I try to put zero here, "it says theta cannot be equal to zero "F of X is not defined at zero." This expression is defined at zero but this tells me, "Well, I really shouldn't apply zero to this function." But we know that if we can find another function that is defined, that is the exact same thing as F of X except at zero, and it is continuous at zero. And so we could say, "G of X is equal to one over two plus two "cosine theta." Well then we know this limit is going to be the exact same thing as the limit of G of X, as theta approaches zero. Once again, these two functions are identical except F of X is not defined at theta equals zero, while G of X is. But the limits as theta approaches zero are going to be the same. And we've seen that in previous videos. And I know what a lot of you are thinking. Sal, this seems like a very, you know, why don't I just, you know, do this algebra here. Cross these things out of this. Substitute zero for theta. Well you could do that and you would get the answer, but you need to be clear if, or it's important to be mathematically clear of what you are doing. If you do that, if you just crossed these two out and all of a sudden you're expression becomes defined at zero, you are now dealing with a different expression or a different function definition. So to be clear, if you want to say this is the function you're finding the limit of, you have to put this constraint in to make sure it has the exact same domain. But lucky for us, we can say, if we've had another, another function that's continuous at that point that doesn't have that gap there, that doesn't have that point discontinuity out, the limits are going to be equivalent. So the limit as theta approaches zero of G of X, well, that's just going to be since it's continuous at zero. We could say that's just going to be, we can just substitute. That's going to be equal to G of zero which is equal to one over two plus cosine two, one over two plus two times cosine of zero. Cosine of zero is one, so it's just one over two plus two, which is equal to, deserve a little bit of a drum roll here. Which is equal to one fourth. And we are done.